If $p(x)$ is a nonconstant polynomial with real coefficients, then how can we show that $$ \lim_{n\to \infty} \int_0^1 e^{i\cdot n \cdot p(x)}~dx=0 ?$$
The integrand $e^{i \cdot n \cdot p(x)}$ is clearly bounded by $1$, but I can't apply the dominated convergence theorem because $\lim_{n\to \infty}e^{i \cdot n \cdot p(x)}$ does not necessarily exists. Any hints?
Given $\delta > 0$, let $[a,b] \subset [0,1]$ be an interval on which $p' > \delta$. Then using the change of variable $t = p(x)$
$$ \int_a^b \exp(inp(x))\; dx = \int_{p(a)}^{p(b)} \frac{\exp(int)}{p'(p^{-1}(t))} \; dt $$ (where $p^{-1}$ is the inverse function to the restriction of $p$ to the interval $[a,b]$) and this converges to $0$ as $n \to \infty$ by the Riemann-Lebesgue Lemma. Similarly for intervals on which $p' < -\delta$. Now take $\epsilon \to 0+$
Given $\epsilon > 0$, after excluding a set of measure $< \epsilon$ containing the zeros of $p'$ we cover the rest by finitely many intervals on which, for some $\delta > 0$, $p' > \delta$ or $p' < -\delta$, and conclude that $$\limsup_{n \to \infty} \left|\int_0^1 \exp(inp(x))\; dx \right| < \epsilon$$